Question: Simplify and expand the following expression: $ \dfrac{2n + 7}{2n + 1}-\dfrac{4n + 6}{n + 9} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2n + 1)(n + 9)$ Multiply the first term by $\dfrac{n + 9}{n + 9}$ $ \begin{align*} \dfrac{2n + 7}{2n + 1} \times \dfrac{n + 9}{n + 9} & = \dfrac{(2n + 7)(n + 9)}{(2n + 1)(n + 9)} \\ & = \dfrac{2n^2 + 25n + 63}{(2n + 1)(n + 9)}\end{align*} $ Multiply the second term by $\dfrac{2n + 1}{2n + 1}$ $ \begin{align*} \dfrac{4n + 6}{n + 9} \times \dfrac{2n + 1}{2n + 1} & = \dfrac{(4n + 6)(2n + 1)}{(n + 9)(2n + 1)} \\ & = \dfrac{8n^2 + 16n + 6}{(n + 9)(2n + 1)}\end{align*} $ Now we have: $ = \dfrac{2n^2 + 25n + 63}{(2n + 1)(n + 9)} - \dfrac{8n^2 + 16n + 6}{(n + 9)(2n + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2n^2 + 25n + 63 - (8n^2 + 16n + 6)}{(2n + 1)(n + 9)} $ $ = \dfrac{2n^2 + 25n + 63 - 8n^2 - 16n - 6}{(2n + 1)(n + 9)} $ $ = \dfrac{-6n^2 + 9n + 57}{(2n + 1)(n + 9)}$ Expand the denominator: $ = \dfrac{-6n^2 + 9n + 57}{2n^2 + 19n + 9}$